Question

Practical 2: • A. Solve the following difference equations using the sequential method: y[k] – y[k – 1] = x[X] + x[k – 1] where x[k] = 0.8ku[k] and y-1) = 0. B. Plot the results for the range o sk s 20. • A. Solve the following difference equations using the z-transform: y[k] + 2.5y[k – 1] +y[k – 2] = x[k] + x[k – 1] where x[k] = 0.8ku[k]. B. Plot the results for the range o sk s 20.

Answer 1

Please feel free to ask doubts in the comment sectio.
Since this not feasible for solving with hands beause finding (0.8)^20 by hand is not a good idea so I guess you wanted the solutions using MATLAB but if you wanted handwritten solutions just tell me in comments and I will upload hand written solutions too.

1.

close all;
clear;
clc;

y = 0;
k = 0:25;
x = (0.8).^k;
for n=2:25
   y(n) = y(n-1) + x(n) +x(n-1);
end

stem(0:20,x(1:21));
xlabel("n");
hold on
stem(0:20,y(1:21));
title("Plot for input and output");
legend("x(n)","y(n)");

Plot for input and output - yon) 8 P 12

2.

close all;
clear;
clc;

syms z n x(n) y(n) X Y;
eqn = y(n) + 2.5*y(n-1)+y(n-2) == x(n) + x(n-1);
eqn_z = ztrans(eqn);
eqn_z = subs(eqn_z,[ztrans(y(n), n, z) ztrans(x(n), n, z)],[X Y]);
eqn_z = subs(eqn_z,[y(-1) y(-2) x(-1)],[0 0 0]);
Y = solve(eqn_z,Y);
H = simplify(Y/X);
disp("The Transfer function in z domain = ");
pretty(H)
assume(n>0);
x(n) = (0.8)^n;
Xx = ztrans(x);
Y = subs(Y,X,Xx);
yn = iztrans(Y);
n_range = 0:20;
stem(n_range,subs(x,n,n_range));
xlabel("n");
hold on;
stem(n_range,subs(yn,n,n_range));
title("Plot for input and output");
legend("x(n)","y(n)");

The Transfer function in z domain = 2 2 2 + 5z + 2 22 (z + 1)

Plot for input and output + - ) yin) 0.5 20

Plot for input and output - yon) 8 P 12

The Transfer function in z domain = 2 2 2 + 5z + 2 22 (z + 1)